STEM Problem Detail

Problem Statement

Prompt

provide me a step by step solution for problem 34.

Solution Rendering...

**Given:**

Temperature,
\[
T = 298\ \text{K}
\]

Pressure,
\[
P = 1.0133 \times 10^{5}\ \text{Pa}
\]

Gases:
\[
\mathrm{CO}, \quad \mathrm{N_2}
\]

Heat capacities at constant pressure:
\[
C_{p,\mathrm{CO}} = 29.15\ \text{J mol}^{-1}\text{K}^{-1}
\]
\[
C_{p,\mathrm{N_2}} = 29.10\ \text{J mol}^{-1}\text{K}^{-1}
\]

Internuclear distances:
\[
r_{\mathrm{CO}} = 1.128 \times 10^{-10}\ \text{m}
\]
\[
r_{\mathrm{N_2}} = 1.098 \times 10^{-10}\ \text{m}
\]

Entropies:
\[
S_{\mathrm{CO}} = 197.4\ \text{J mol}^{-1}\text{K}^{-1}
\]
\[
S_{\mathrm{N_2}} = 191.5\ \text{J mol}^{-1}\text{K}^{-1}
\]

**Theory:**

For an ideal diatomic gas, the molar entropy is given by
\[
S = S_{\text{trans}} + S_{\text{rot}} + S_{\text{vib}} + S_{\text{elec}}
\]

At ordinary temperatures, vibrational and electronic contributions are negligible:
\[
S \approx S_{\text{trans}} + S_{\text{rot}}
\]

**1. Translational entropy:**

The translational entropy of an ideal gas is
\[
S_{\text{trans}} =
R \left[
\ln \left(
\frac{(2 \pi m k T)^{3/2}}{h^3}
\frac{kT}{P}
\right)
+ \frac{5}{2}
\right]
\]

Since CO and N$$_2$$ have **equal molecular weights**,
\[
S_{\text{trans}}(\mathrm{CO}) = S_{\text{trans}}(\mathrm{N_2})
\]

**2. Rotational entropy:**

For a linear molecule,
\[
S_{\text{rot}} =
R \left[
\ln \left( \frac{T}{\sigma \theta_r} \right) + 1
\right]
\]

where
\[
\theta_r = \frac{h^2}{8 \pi^2 I k}
\]
and
\[
I = \mu r^2
\]

**Symmetry numbers:**
\[
\sigma_{\mathrm{CO}} = 1
\]
\[
\sigma_{\mathrm{N_2}} = 2
\]

**Moments of inertia:**

Because the bond length of CO is larger,
\[
I_{\mathrm{CO}} > I_{\mathrm{N_2}}
\]

Thus,
\[
\theta_{r,\mathrm{CO}} < \theta_{r,\mathrm{N_2}}
\]

**Comparison of rotational entropy:**

The rotational entropy increases when:

- symmetry number $$\sigma$$ decreases
- moment of inertia $$I$$ increases

CO has:

- lower symmetry ($$\sigma = 1$$)
- larger bond length and moment of inertia

Therefore,
\[
S_{\text{rot}}(\mathrm{CO}) > S_{\text{rot}}(\mathrm{N_2})
\]

**Explanation of entropy difference:**

Since translational contributions are identical and vibrational contributions are negligible,
\[
S_{\mathrm{CO}} - S_{\mathrm{N_2}}
\approx
S_{\text{rot}}(\mathrm{CO}) - S_{\text{rot}}(\mathrm{N_2})
\]

The lower symmetry and larger moment of inertia of CO lead to a significantly larger rotational entropy.

**Final Explanation:**

\[
\boxed{
\text{The higher entropy of CO compared to } \mathrm{N_2} \text{ arises from its lower symmetry number}
}
\]

\[
\boxed{
\text{and its larger moment of inertia, which increase the rotational entropy.}
}
\]

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Problem #53